Limit Comparison Test
The Limit Comparison Test states that if
\sum a_n
and
\sum b_n
are series with positive terms, then:
If
\lim_{n \to \infty} a_n/b_n = c
converges, then
\sum a_n
and
\sum b_n
either both converge or both diverge.
Examples
Example 1: Convergence using Direct Comparison Test
Determine whether the series
\sum_{n=1}^{\infty} \frac{1}{n^2 + 1}
converges or diverges.
\begin{align*}
\text{For all } n &\geq 1, \\
\quad n^2 + 1 &> n^2 \\
\frac{1}{n^2 + 1} &< \frac{1}{n^2}
\end{align*}
We know that
\sum_{n=1}^{\infty} \frac{1}{n^2}
is a p-series with
p = 2 > 1
, so it converges.
Since
0 < \frac{1}{n^2 + 1} < \frac{1}{n^2}
for all
n \geq 1
, by the Direct Comparison Test, the series
\sum_{n=1}^{\infty} \frac{1}{n^2 + 1}
also converges.
Example 2: Divergence using Direct Comparison Test
Determine whether the series
\sum_{n=1}^{\infty} \frac{1}{\sqrt{n} - 0.5}
converges or diverges.
\begin{align*}
\text{For } n &\geq 1, \\
\sqrt{n} &> \sqrt{n} - 0.5 \\
\frac{1}{\sqrt{n}} &< \frac{1}{\sqrt{n} - 0.5}
\end{align*}
We know that
\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}} = \sum_{n=1}^{\infty} \frac{1}{n^{1/2}}
is a p-series with
p = \frac{1}{2} < 1
, so it diverges.
Since
\frac{1}{\sqrt{n} - 0.5} > \frac{1}{\sqrt{n}} > 0
for all
n \geq 1
, by the Direct Comparison Test, the series
\sum_{n=1}^{\infty} \frac{1}{\sqrt{n} - 0.5}
also diverges.
Example 3: Convergence using Limit Comparison Test
Determine whether the series
\sum_{n=1}^{\infty} \frac{3n - 2}{n^3 + 4n + 1}
converges or diverges.
\begin{align*}
\text{Let } a_n &= \frac{3n - 2}{n^3 + 4n + 1} \\
\text{Let } b_n &= \frac{1}{n^2}
\end{align*}
Consider the limit:
\begin{align*}
\lim_{n \to \infty} \frac{a_n}{b_n} &= \lim_{n \to \infty} \frac{\frac{3n - 2}{n^3 + 4n + 1}}{\frac{1}{n^2}} \\
&= \lim_{n \to \infty} \frac{n^2(3n - 2)}{n^3 + 4n + 1} \\
&= \lim_{n \to \infty} \frac{3n^3 - 2n^2}{n^3 + 4n + 1} \\
&= \lim_{n \to \infty} \frac{3 - \frac{2}{n}}{1 + \frac{4}{n^2} + \frac{1}{n^3}} = 3
\end{align*}
Since
\lim_{n \to \infty} \frac{a_n}{b_n} = 3
is a finite positive number, and
\sum_{n=1}^{\infty} \frac{1}{n^2}
converges (p-series with
p = 2 > 1
),
by the Limit Comparison Test, the series
\sum_{n=1}^{\infty} \frac{3n - 2}{n^3 + 4n + 1}
also converges.